Is there any subsequence of a funtion sequence converges at every point Let $f_n(x)＝\sin{nx}，n＝1,2,…$ be a sequence of functions on $\mathbb{R}$.
Is there a subsequence
$f_{n_j}(x)$ that converges at every point $x \in \mathbb{R}$？
(Note that the subsequence must not rely on the point $x$.)
Thanks in advance!
 A: Let $f_n(x)$ be the $n$-th digit in the binary expansion of $x$. If you equipp $[0,1]$ with the Lebesgue measure, it is easy to check that $(f_n)_n$ are independent random variables and $\lambda(f_n=0)=\lambda(f_n=1)=1/2$.
Clearly, any subsequence $f_{n_k}$ is again a sequence of independet random variables.
Let $(f_{n_k})_k$ be an arbitrary subsequence of $(f_n)_n$. The strong law of large number implies that there is a set $A \subset [0,1]$ with $\lambda(A)=1$ s.t.
$$
\lim_{N \to \infty} \frac{1}{N} \sum_{k=1}^N f_{n_k}(x) \to 1/2
$$
for all $x \in A$.
Fix $x$: If $(f_{n_k}(x))_k$ converges, then its limit can only be $0$ or $1$. In this case, its Cesaro mean is convergent as well with the same limit. Hence, for all $x \in A$ the $(f_{n_k}(x))_k$ is not convergent. Since $A$ has measure 1, it is in particular nonempty. Therefore, $(f_{n_k})_k$ dosn't converge pointwise.
Remark: Since $[0,1]^{[0,1]}$ is compact by Tychonoff's theorem, that example shows that compactness doesn't imply sequential compactness.
